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Topic: Lowering voltage by a specific amount? (Read 611 times) previous topic - next topic

patrickc01

I bought a sparkfun laser card (http://www.sparkfun.com/datasheets/Components/Laser-Card.pdf)
It states that the voltage should be 3.1 +/- 10%.  I was trying to figure out how I can lower the voltage from 3.5 (or 5V) to 3.1.  Is there a way to adjust voltage like this?  Or would it be ok to put a diode in the circuit (from what I understand this would lower the voltage by ~.7).
Thanks for any help!

majenko

It all depends on how much current draw you're looking at, and how precise you want to get the voltage.

You can use an adjustable "Low Drop-Out" voltage regulator to get a precise voltage from the 5v connection.  This would be the most precise way.

Yes, a diode will give a roughly 0.7v drop per diode.  You can't get 3.1v using silicon diodes alone.  A combination of silicon and schottky diodes could be employed, but that would probably cost more than an LDO regulator.

If the current draw is a constant value (say for an LED), then a simple resistor can be used to drop the voltage.  Using Ohm's law you can calculate the value of the resistor to use.
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patrickc01

Thanks for the answer.  I was under the impression that resistors lower current, not voltage, guess I was wrong..  Havent heard about schottky diodes, putting it on my list of things to learn

Jack Christensen


Thanks for the answer.  I was under the impression that resistors lower current, not voltage, guess I was wrong..  Havent heard about schottky diodes, putting it on my list of things to learn


Resistance is the constant of proportionality between voltage and current. See http://en.wikipedia.org/wiki/Ohms_law
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Techylah

Use a germanium diode, like a 1N34A, instead of silicon.   It has a drop of .15 to .3v depending on current.  Perhaps use two.

Magician

Resistor should be o'k for laser, knowing current from the spec. data, do a math : R = ( 5 - 3.1 ) / 35 mA = 54.28 Ohm

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